HELPSubstitution and Integral by Parts

In summary, the Rayleigh distribution has a probability density function that is given by f(x) = (x/a)*e-x2/(2a), for x>0. The question asks for the expected value of x2 for a random sample from this distribution.
  • #1
moumouer
9
0
I'm having big trouble when trying to figure this integral out. Please help!

Integral (from 0 to infinity): ((x^2)/a)*e^[(-x^2)/2a] dx a is a constantThanks in advance!
 
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  • #2
I would start with u = x and dv = (1/a)x*e-x2/(2a)dx.
 
  • #3
OHHHH! Thank you sooo much! I'v been thinking my head off by trying to do substitution!

And I have one more question, I'm required to calculate E(x2) too, so now instead of (X^2)/a for the first part, it becomes (x^3)/a, and the 2nd part remains the same. How do I approach this one?

Thank you very much!
 
  • #4
Oh wait a minute, after doing the parts, since V = - (e-x2/(2a)), now applying the fomula, uv - integral v du, how do I solve Integral - (e-x2/(2a)) dx ?
 
  • #5
Hi Mark44, I think your method works perfectly w/ my 2nd question. I worked it out already. But I still don't know how to solve the 1st question.
 
  • #6
Then I think you're stuck, unless there is some additional information we haven't seen yet. e-x2 doesn't have a nice neat antiderivative.
 
  • #7
The question says Let X1...Xn be a random sample from a Rayleigh distribution with pdf
f(x) = (x/a)*e-x2/(2a), for x>0. That's it! And I'm suppose to find E(x).
 
  • #9
thank you so much!
 
  • #10
Hi Dick, I read through the article on wiki, and I'm wondering how the parameter a in my problem affect the integral. I cannot ignore a here since the reason I'm calculating E(x) is becoz I'm trying to find the Method of Moment estimator of a.

Thank you!
 
  • #11
Set u^2=x^2/(2a). So u=x/sqrt(2a). If you write the integral in terms of u, you should be able to collect all of the a's outside of the integral.
 
  • #12
But it still doesn't solve the integral. it became: Integral 2u^2*e^(-u2)sqrt(2a)du
 
  • #13
Isn't it 2*sqrt(2a)*integral u^2*e^(-u^2)*du? The u integral is just a constant. I thought you were going to use integration by parts like Mark44 suggested to reduce it to the Gaussian integral?
 
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  • #14
Thank you very much!
 

Related to HELPSubstitution and Integral by Parts

1. What is substitution in calculus?

Substitution in calculus is a technique used to simplify complex integrals or derivatives by substituting a new variable or expression in place of the original one. This allows for easier integration or differentiation.

2. How do you perform substitution in calculus?

To perform substitution in calculus, you first identify a part of the integral or derivative that can be substituted with a new variable. Then, you choose an appropriate substitution that will simplify the expression. Finally, you substitute the new variable in the integral or derivative and solve for the final result.

3. What is the purpose of substitution in calculus?

The purpose of substitution in calculus is to simplify complex integrals or derivatives that are difficult to compute. It allows for easier integration or differentiation by substituting a new variable or expression in place of the original one.

4. How is substitution related to the chain rule?

Substitution in calculus is closely related to the chain rule. The chain rule states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. Substitution can be seen as a way to reverse this process, by replacing the inner function with a new variable and integrating or differentiating the outer function.

5. Can substitution be used in both integrals and derivatives?

Yes, substitution can be used in both integrals and derivatives. In integrals, substitution is used to simplify the integration process, while in derivatives, it is used to simplify the differentiation process. However, the specific method of substitution may differ slightly between integrals and derivatives.

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